anonymous
  • anonymous
(2x^2+5x-12)/(x^2-4x)<0
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
Must be 6X
anonymous
  • anonymous
show work?
anonymous
  • anonymous
factor the numerator and denominator: [(2x-3)*(x+4)] / [x*(x-4)] ....hmm nothing cancels... btw, what do you want done here?

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More answers

anonymous
  • anonymous
It is 6x
anonymous
  • anonymous
i think he needs to solve for x. move the denominator to the other side of the equation, then start solving by using algebra. The rule to remember when dealing with < or > signs is to FLIP THE SIGN each time you are multiplying or dividing by a negative number.
anonymous
  • anonymous
yea it is solving for x. although i got to the factoring but i was unsure of what to do next
anonymous
  • anonymous
6x
anonymous
  • anonymous
ok, then we need to solve, that is, find all the x-values that will make the stament true...
anonymous
  • anonymous
so after multiplying and doing algebra, you should have x^2 +9x + - 12 <0
anonymous
  • anonymous
from the factored expression [(2x-3)*(x+4)] / [x*(x-4)] find all x values that will make the numerator = 0 then all x values that will make the denominator = 0
anonymous
  • anonymous
for numerator: x = 3/2, -4 for denominator: x = 0, 4 now pick values within those interval and check what will make f positive or negative: |dw:1332627744298:dw|
anonymous
  • anonymous
-4
anonymous
  • anonymous
hello? not here anymore...
anonymous
  • anonymous
trying to figure this out... this is confusing me
anonymous
  • anonymous
are u there?
anonymous
  • anonymous
yea.. i dont get it still
anonymous
  • anonymous
do you understand how I factored to get the new expression: [(2x-3)*(x+4)] / [x*(x-4)] < 0
anonymous
  • anonymous
yes
anonymous
  • anonymous
ok, do you understand @kumar2006 answers? where do you think he got those numbers?
anonymous
  • anonymous
oh, isee. is that the final answer?
anonymous
  • anonymous
kumar2006 answers are correct. but do you see how he got those solution sets?
anonymous
  • anonymous
yes, final answer.
anonymous
  • anonymous
alright i get it now, thanks
anonymous
  • anonymous
yw

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